Question

Cool solution.

Answer 1

Problem:

Prove that\[\left(\frac{1 + a}{1+b}\right)^{\frac{1}{a-b}}< e\]for distinct reals \(a,b\).

Answer 2

How many seconds can I have to think about it before you give the solution? :p

Answer 3

Try it out. Just meant to share this problem.

Answer 4

\[(\frac{1+a}{1+b})^{\frac{1}{a-b}} \\ =(\frac{1+b+a-b}{1+b})^\frac{1}{a-b} \\ =(1+\frac{a-b}{1+b})^{\frac{1}{a-b}}\]
\[=(1+\frac{1}{\frac{1+b}{a-b}})^\frac{1}{a-b} \\ =(1+\frac{\frac{1}{1+b}}{\frac{1}{a-b}})^\frac{1}{a-b}\]

\[\lim_{a-b \rightarrow \infty}(1+\frac{\frac{1}{1+b}}{\frac{1}{a-b}})^{\frac{1}{a-b}}=\lim_{a-b \rightarrow \infty} e^{\frac{1}{1+b}} \\ ...\]
if a-b goes to infinity
I don't think we can say anything about the limit of b from that
so that limit should be just exp(1/(1+b))
..I don't know if this is the right route
For some reason this is the first thing that popped in my head

Answer 5

Ganeshie is here - he'll easily crack this one. :|

Answer 6

I did something wrong

Answer 7

I should have said 1/(a-b) goes to infinity

Answer 8

which would mean a-b goes to 0

Answer 9

`ganeshie8 is typing a reply...`

Well, here we go.

Answer 10

$$\left(\frac{1 + a}{1+b}\right)^{\frac{1}{a-b}}< e\\\frac1{a-b}\log\left(\frac{1+a}{1+b}\right)<1\\\frac1{a-b}\left(\log(1+a)-\log(1+b)\right)<1\\\log(1+a)-\log(1+b)<(1+a)-(1+b)$$... is just a restatement of the fact that \(\log\) is concave by Jensen's inequality

Answer 11

all those steps are reversible, so the proof just follows from the fact that \(\log\) is concave via Jensen's inequality

Answer 12

I don't think my way is good enough it only shows for a close to b
we have
exp(1/(1+b))
but I think this right here is <e for any real b

Answer 13

\[e^{x-y} \gt 1+(x-y)+(x-y)^2/2 \gt 1+x-y = 1+\dfrac{x-y}{1+y} = \frac{1+x}{1+y}\]

Answer 14

haha that one is great

Answer 15

but hmm @ganeshie8 both you and i have only proven it for \(a-b>0\) technically

Answer 16

Ah right, I see..

Answer 17

unless you can rigorously show that the even terms outweigh the odd ones in the expansion \(e^x=1+x+\frac12 x^2+\dots\)

Answer 18

https://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_for_the_remainder

Answer 19

looks the taylor series one also requires \(y\gt 0\)

\[e^{x-y} \gt 1+(x-y)\gt 1+\dfrac{x-y}{1+y} = \frac{1+x}{1+y}\]

Answer 20

@parthkohli share your answer

Answer 21

@ganeshie8 That's exactly my answer. I think I got the incomplete question.

Answer 22

Hey, your next project is to implement LaTeX in the drawing tool. That looks good.

Answer 23

looks the inequality is true only when \(a,b\) have same signs